The roots of the eqn x^2-8x+h
are (alpha) and (alpha+3k)
Express h in terms of k
For
Quadratic equation
ax^2 +bx +c =0
Sum of roots of quadratic equation $\displaystyle = \frac{-b}{a}$
Product of roots of equation $\displaystyle = \frac{c}{a}.....(2) $
Hence using (2) we get
$\displaystyle
(\alpha) \times (\alpha + 3k) = h --------(3)$
And
$\displaystyle (\alpha) + (\alpha+3k) = 8 $
$\displaystyle \alpha= (8-3k)/2 --------------(1) $
Use one in (3)
$\displaystyle \frac{(8-3k)}{2} \times \frac{8+3k}{2} = h $
$\displaystyle 64- 9k^2 = 4h$
I will guess that the author of the assignment forgot the "equals zero" part of the "equation". (Without the "equals" sign, it's not an "equation"; without the assumption that the other side was "zero", there's no way to find the roots.)
In case you haven't learned about the formulas relating the roots of a quadratic function to the coefficients, you can instead use the relationship between roots and factors to find the factorization:
. . . . .$\displaystyle \mbox{If }\, x\, =\, a\, \mbox{ is a root, then }\, x\, -\, a\, \mbox{ is a factor.}$
In your case, the zeroes are:
. . . . .$\displaystyle x\, =\, \alpha\, \mbox{ and }\, x\, =\, \alpha\, +\, 3k$
Then the factors must have been:
. . . . .$\displaystyle x\, -\, \alpha\, \mbox{ and }\, x\, -\, (\alpha\, +\, 3k)$
Multiplying, we get:
. . . . .$\displaystyle (x\, -\, \alpha)(x\, -\, (\alpha\, +\, 3k))\, =\, x^2\, +\, (-2\alpha\, -\, 3k)x\, +\, (-\alpha^2\, -\, 3\alpha k)$
Comparing coefficients, we see that:
. . . . .$\displaystyle -2\alpha\, -\, 3k\, =\, -8$
Then:
. . . . .$\displaystyle \alpha\, =\, 4\, -\, \frac{3}{2}k$
Comparing coefficients again, we see that:
. . . . .$\displaystyle -\alpha^2\, -\, 3\alpha k\, =\, h$
Substituting from above, we get:
. . . . .$\displaystyle -\left(4\, -\, \frac{3}{2}k\right)^2\, -\, 3\left(4\, -\, \frac{3}{2}k\right)k\, =\, h$
. . . . .$\displaystyle -16\, +\, 12k\, -\, \frac{9}{4}k^2\, -\, 12k\, +\, \frac{9}{2}k^2\, =\, h$
...which simplifies to give a result equivalent to the previous reply.